# How to derive the short run cost function

Given the production function $$f(K, L)=\min\{3K,2L\}$$, the procedure to find the long-run cost function would be to use the condition: $$3K=2L=Y$$ where $$K=\frac{\overline{Y}}{3}$$ and $$L=\frac{\overline{Y}}{2}$$. $$K$$ is units of capital; $$L$$ is units of labor; $$Y$$ is units of output.

Now, I am being asked to find the short-run cost function where labor is fixed at $$L>\frac{\overline{Y}}{2}$$. I am not sure how to approach the problem in this case.

• Since you say labor is fixed, I guess you need the short run cost function where $K$ is the factor that represents the variable cost, while $L$ representing the fixed cost. Commented May 23, 2023 at 13:39
• Don't forget you can accept an answer by clicking on the checkmark if you found it helpful! Commented May 29, 2023 at 20:20

The long run is defined as the time period in which all the inputs can be changed as per the firm's desire i.e., for a production function of the form $$f(K,L)$$ both $$K$$ and $$L$$ are variable in determining output.

The long-run cost minimization problem is: \begin{align} \min_{K,L} \quad &wL+rK \\ \textrm{s.t.} \quad& \min(3K,2L)\geq \bar Y \end{align}

The constraint binds at optimum and solving the optimization problem yields conditional input demands $$K^d(w,r,\bar Y)=\frac{\bar Y}{3}$$ and $$L^d(w,r,\bar Y)=\frac{\bar Y}{2}$$. Therefore, LR Cost function is: $$C(w,r,\bar Y)=\bar Y(\frac{w}{2}+\frac{r}{3})$$

The short run is defined as that time period in which one or more inputs are fixed.

The short-run cost minimization problem is: \begin{align} \min_{K} \quad & w \bar L+r K \\ \textrm{s.t.} \quad & \min(3K,2 \bar L)\geq\bar Y \\ \end{align}

for the constraint $$\min(3K,2\bar L)\geq \bar Y$$ to hold we need $$3K\geq \bar Y$$ and $$2\bar L\geq \bar Y$$

therefore we can rewrite the problem as: \begin{align} \min_{K \geq \frac{\bar{Y}}{3}} \quad & w \bar L+r K \\ \end{align}

notice that $$w\bar L+rK$$ is increasing in $$K$$ therefore in order to solve the above problem we set $$K$$ to the lowest value it can take.
Thus, $$K^d_{SR}(w,r,\bar L, \bar Y)=\frac{\bar Y}{3}$$ and consequently SR cost function is: $$C_{SR}(w,r,\bar L, \bar Y)=w\bar L+r\frac{\bar Y}{3}$$

• In case we were given that $\bar L<\frac{\bar Y}{2}$ in that case to minimize cost, we would set $3K=2\bar L <\bar Y$ i.e., $K_{SR}^d(w,r,\bar L, \bar Y)=\frac{2\bar L}{3}$. So, I think a more complete solution to the SR cost minimization problem is: $K_{SR}^d(w,r,\bar L, \bar Y)=\begin{cases} \frac{\bar Y}{3} & \text{if } \bar L>\frac{\bar Y}{2} \\ \frac{2 \bar L}{3} & \text{if } \bar L\leq \frac{\bar Y}{2} \end{cases}$ but I'm not sure how correct this is, so decided to include this in the comments Commented May 24, 2023 at 10:27
• Thank you! I'm also not sure about the second part but thanks anyway! Commented May 29, 2023 at 19:21

Since $$L$$ is fixed at some $$\overline{L} > \frac{\overline{Y}}{2}$$, then $$\overline{Y} < 2L$$.

This implies the $$\min$$ term equals $$3K$$.

Therefore, $$f(K,\overline{L}) = 3K$$.

With this, the problem we want to solve is

$$\min_K w\overline{L} + rK$$

s.t. $$3K \geq \overline{Y}$$

Solving for $$K$$ we get $$K \geq \frac{\overline{Y}}{3}$$

Clearly the objective function is minimized at equality, so we get $$K = \frac{\overline{Y}}{3}$$

Plugging into the objective function, the short run minimum cost function is

$$C(r,w,\overline{Y};\overline{L}) = w \overline{L} + \frac{r \overline{Y}}{3}$$